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Geometry Level 3

A lattice cube is a cube in 3-dimensional space, whose vertices are lattice points. Describe the set of possible side lengths of a lattice cube.

Note: A lattice point is a point whose coordinates are all integers. ( 0 , 0 , 1 ) (0,0,1) is a lattice point, but ( 1 2 , 2 , 0.2 ) (\frac{1}{2}, \sqrt{2}, -0.2 ) is not a lattice point.

For similar problems, you can read my note on Construction .

{ n n is the sum of 3 perfect squares. } \{\sqrt{n}\mid n \text{ is the sum of 3 perfect squares.}\} { n n is a non-negative integer. } \{ n \mid n \text{ is a non-negative integer.} \} { n n is a non-negative integer. } \{ \sqrt{n} \mid n \text{ is a non-negative integer.} \} { n 3 n is a non-negative integer. } \{ \sqrt[3]{n} \mid n \text{ is a non-negative integer.} \}

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2 solutions

Calvin Lin Staff
Mar 3, 2014

For most people, they will think "The side length must be from a lattice point, to another lattice point. Hence, it must be of the form x 2 + y 2 + z 2 \sqrt{x^2 + y^2 + z^2 } ." This leads them to choose { n n is the sum of 3 perfect squares. } \{ n \mid n \text{ is the sum of 3 perfect squares.} \} , which is not correct.

As this problem is on Construction , we need to show that such a lattice cube with those side lengths actually exists. This may not always be true. For example, let's show that there is no lattice cube with side length 3 = 1 + 1 + 1 \sqrt{3} = \sqrt{1 + 1 + 1 } .

Proof: WLOG, one of the vertices is at ( 0 , 0 , 0 ) (0,0,0) . Then, the 3 vertices that are connected to it, are of the form ( ± 1 , ± 1 , ± 1 ) (\pm 1, \pm 1, \pm 1 ) . Observe that the distance between any 2 of these vertices is either 2, 8 \sqrt{8} or 12 \sqrt{12} . However, if they were the vertices of a cube of side length 3 \sqrt{3} , then the distance must be 3 × 2 \sqrt{3} \times \sqrt{2} . Hence, we have a contradiction. _ \square

This shows us that the question is actually more complex and subtle. There are several ways to proceed, and I'd present a proof using vectors, as that is slightly more approachable than the proof using matrices.

Proof (general case): Let the cube have side length k \sqrt{k} . (Why must it be in this form?) Let one of the vertices be ( 0 , 0 , 0 ) (0,0,0) . Let the 3 adjacent vertices be of the form ( a x , a y , a x ) , ( b x , b y , b z ) , ( c x , c y , c z ) (a_x, a_y, a_x), (b_x, b_y, b_z), (c_x, c_y, c_z) . Then, we know that

( a x a y a z ) × ( b x b y b z ) = ± k ( c x c y c z ) \begin{pmatrix} a_x \\ a_y \\ a_z \end{pmatrix} \times \begin{pmatrix} b_x \\ b_y \\ b_z \end{pmatrix} = \pm \sqrt{k} \cdot \begin{pmatrix} c_x \\ c_y \\ c_z \end{pmatrix}

Since the LHS are all integers, hence the RHS are also all integers. This implies that k \sqrt{k} must be an integer.

Conversely, for any integer n n , there is clearly a lattice cube of side length n n . Hence, the set of side lengths of a lattice cube is { n n is an integer. } \{ n \mid n \text{ is an integer.} \} . _ \square

I am sorry, but I cannot get you.I am not even sure what you are implying when you say side length.My experience tells me that it is the length of the edge of the cube and can have any possible value and not just integral values.Do lengths of lattice cubes have integral constraints on them.Is it because lengths are being calculated with respect to an arbitrary measure that says that side lengths are measured in real number multiples of the smallest actual length of the cube(I am guessing there is something like this or the question would be ordinary).I am not sure I am even making sense but I request you to try and understand my dilemma.

A Former Brilliant Member - 7 years, 3 months ago

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A lattice point is a point whose coordinates are all integers. ( 0 , 0 , 1 ) (0,0,1) is a lattice point, but ( 1 2 , 2 , 0.2 ) (\frac{1}{2}, \sqrt{2}, -0.2 ) is not a lattice point.

Hence, the minimum non-zero distance between 2 lattice points is 1, and so the minimum non-zero side length of a lattice cube is 1.

Calvin Lin Staff - 7 years, 3 months ago

what about n being negative?

jaismin kaur - 7 years, 3 months ago

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I'd edit the options to make them non-negative.

Calvin Lin Staff - 7 years, 3 months ago
Prem Kumar
Mar 24, 2014

since the edges have to be integral, a non integral length for the side of the cube would result into at least one of the edges to have a non-integral co-ordinate... Hence, the sides have to be a positive integer...

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